已知数列an满足a1+2a2+22a3+…+2n-1an=n2(n∈N*).(Ⅰ)求数列{an}的通项;(Ⅱ)若bn=nan求数列{bn}的前n项Sn和.-数学

题目简介

已知数列an满足a1+2a2+22a3+…+2n-1an=n2(n∈N*).(Ⅰ)求数列{an}的通项;(Ⅱ)若bn=nan求数列{bn}的前n项Sn和.-数学

题目详情

已知数列an满足a1+2a2+22a3+…+2n-1an=
n
2
(n∈N*).
(Ⅰ)求数列{an}的通项;
(Ⅱ)若bn=
n
an
求数列{bn}的前n项Sn和.
题型:解答题难度:中档来源:不详

答案

(Ⅰ)n=1时,a1=class="stub"1
2

a1+2a2+2a3…+2n-1an=class="stub"n
2
…..(1)
∴n≥2时,a1+2a2+2a3…+2n-2an-1=
n-1
2
….(2)
(1)-(2)得2n-1an=class="stub"1
2
an=class="stub"1
2n

a1=class="stub"1
2
也适合上式,∴an=class="stub"1
2n

(Ⅱ)bn=n•2n,∴Sn=1•2+2•22+3•23+…+n•2n(3)
2Sn=1•22+2•23+…+(n-1)•2n+n•2n+1(4)
(3)-(4)可得-Sn=1•2+1•22+1•23+…+1•2n-n•2n+1
=
2(1-2n)
1-2
-n•2n+1=2n+1-2-n•2n+1

Sn=(n-1)•2n+1+2

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